New answers tagged finite-fields
0
votes
Primitive polynomial for non-prime $\mathbb{F}_q$
If $q$ is not too large, there is a simple series of tests. Find all the prime factors of $q-1$, and then raise a candidate primitive element to all combinations of those prime factors except all of ...
2
votes
Primitive polynomial for non-prime $\mathbb{F}_q$
Short answer: if we can factor polynomials over non-prime fields, then we can find primitive polynomials for them using primitive polynomials for prime fields.
Let $q=p^k$; we want to find a primitive ...
2
votes
Accepted
Foundations of Galois theory
To have $K$ be the fixed points of a subgroup of ${\rm Aut}(L/K)$, that subgroup is ${\rm Aut}(L/K)$ and $L/K$ is a Galois extension.
Note I don't need to assume $L$ is the splitting field of a ...
0
votes
Field $\mathbb{Q}(\alpha)$ with $\alpha=\sqrt[3]7+2i$
For 2 you can use some linear algebra.
Take the target basis $1,\alpha,\alpha ^2,\alpha^3,\alpha^4,\alpha^5$.
Create the matrix with rows designating elements in the basis and columns designating the ...
3
votes
Accepted
Splitting field of $t^6+t^5+t^4+t^3+1$ over $\mathbb{F}_2[t]$
The easy answer relies on the facts that there exists exactly one finite extension of $\mathbb F_2$ of each degree $d\in\mathbb N_+$, denoted $\mathbb F_{2^d}$, and that $\mathbb F_{2^n}\subseteq\...
1
vote
Accepted
Over reducibility of $x^{{q}^{n}} -x+a$ over $\mathbb{F}_q$
Recycling the argument from an earlier answer, as explaining a minor modification in a comment is apparently now against the rules.
Let $\Omega=\overline{\Bbb{F}_q}$ be an algebraic closure, and $F:\...
Community wiki
2
votes
Accepted
A question on cyclotomic extension of finite field.
Let's be a bit careful here. The claim is actually false as stated. The fact that $\zeta\in\Bbb{F}_{q^2}\setminus\Bbb{F}_q$ only implies that $\zeta+\zeta^q\in\Bbb{F}_q$. There is no reason to think ...
Community wiki
0
votes
Group representation over finite fields
Let $G$ be a finite group and let $L/K$ be an extension of finite fields and let $V$ be an absolutely irreducible $L[G]$-module and suppose that $K$ contains the character values of $V$. Let $\...
3
votes
Conjugate to transpose via symmetric matrix
The required symmetric transition matrix $S$ always exists. See the list of references given in the first paragraph of Olga Taussky, The Role of Symmetric Matrices in the Study of General Matrices, ...
Community wiki
3
votes
Accepted
Rational canonical form of a matrix over finite fields
Using the Frobenius decomposition, you can find a counterexample with:
$$A=\begin{pmatrix}
C_{p^2}\\
& C_{p^2}\\
&&C_{p^3}\\
&&&C_q
\end{pmatrix}
\qquad
\text{and}
\qquad
B=\...
3
votes
Rational canonical form of a matrix over finite fields
$\newcommand{rk}{\operatorname{rk}}$Call $p_X$ the characteristic polynomial of a matrix $X$, and $\mu_X$ the minimal polynomial. Call $X\oplus Y$ the block matrix $\begin{bmatrix}X &0\\ 0 &Y\...
Community wiki
3
votes
Can I find the inverse of a polynomial directly in its NTT form?
If $\ g(x)\ $ is the inverse $\mod{x^n-1}\ $ of $\ f(x)\ $ over whatever field $\ F\ $ you're working in, and $\ \omega\ $ is a primitive $\ n^\text{th}\ $ root of unity (in the splitting field of $\ ...
2
votes
Irreducible representations of inhomogeneous linear transformations over $\mathbb{F}_q$
Choose a basis $\mathcal{B}=\{f_a-f_0\}_{a\neq0}$ for $V$, where $f_a$ is defined as follows: $f_a(a)=1$ and $f_a(b)=0$ for every $b\neq a$.
Suppose $W$ is a $G$-subspace of $V$. First, note that $(u,...
0
votes
Number of solutions $x^2-dy^2=b$ in finite field $F$ with $char~F\neq 2$
When there is a single solution to this equation, it
being a curve of degree 2, using the birational correspondence between a line and this curve by projection can get more solutions. A projective ...
2
votes
Accepted
Trace and Norm of an element $\alpha \in \mathbb{F}_{q^{n}}$ is equal to Trace and Determinant of Linear Transform $ T(\beta) = \alpha \beta$
Yes, you indeed need to show that the characteristic polynomial of $T$ is equal $(x - \alpha)(x - \alpha^q)\cdots (x - \alpha^{q^{n-1}})$.
For the basic case where $\Bbb F_{q^n} = \Bbb F_q(\alpha)$, ...
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